Sort the given Array as per given conditions
Last Updated : 20 Feb, 2023
Given an array arr[] consisting of N positive integers, the task is to sort the array such that -
- All even numbers must come before all odd numbers.
- All even numbers that are divisible by 5 must come first than even numbers not divisible by 5.
- If two even numbers are divisible by 5 then the number having a greater value will come first
- If two even numbers were not divisible by 5 then the number having a greater index in the array will come first.
- All odd numbers must come in relative order as they are present in the array.
Examples:
Input: arr[] = {5, 10, 30, 7}
Output: 30 10 5 7
Explanation: Even numbers = [10, 30]. Odd numbers = [5, 7]. After sorting of even numbers, even numbers = [30, 10] as both 10 and 30 divisible by 5 but 30 has a larger value so it will come before 10.
After sorting A = [30, 10, 5, 7] as all even numbers must come before all odd numbers.
Approach: This problem can be solved by using sorting. Follow the steps below to solve the problem:
- Create three vectors say v1, v2, and v3 where v1 stores the numbers which are even and divisible by 5, v2 stores the numbers which are even but not divisible by 5, and v3 stores the numbers which are odd.
- Iterate in the range [0, N-1] using the variable i and perform the following steps-
- If arr[i]%2 = 0 and arr[i]%5=0, then insert arr[i] in v1.
- If arr[i]%2 = 0 and arr[i]%5 != 0, then insert arr[i] in v2.
- If arr[i]%2 = 1 then insert arr[i] in v3.
- Sort the vector v1 in descending order.
- After performing the above steps, print vector v1 then print vector v2 in reverse order, and then v3 as the answer.
Below is the implementation of the above approach:
C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to sort array in the way // mentioned above void AwesomeSort(vector<int> m, int n) { // Create three vectors vector<int> v1, v2, v3; int i; // Traverse through the elements // of the array for (i = 0; i < n; i++) { // If elements are even and // divisible by 10 if (m[i] % 10 == 0) v1.push_back(m[i]); // If number is even but not divisible // by 5 else if (m[i] % 2 == 0) v2.push_back(m[i]); else // If number is odd v3.push_back(m[i]); } // Sort v1 in descending order sort(v1.begin(), v1.end(), greater<int>()); for (int i = 0; i < v1.size(); i++) { cout << v1[i] << " "; } for (int i = v2.size()-1; i >= 0; i--) { cout << v2[i] << " "; } for (int i = 0; i < v3.size(); i++) { cout << v3[i] << " "; } } // Driver Code int main() { // Given Input vector<int> arr{ 5, 10, 30, 7 }; int N = arr.size(); // FunctionCall AwesomeSort(arr, N); return 0; } // Java program for the above approach import java.util.Collections; import java.util.Vector; public class GFG_JAVA { // Function to sort array in the way // mentioned above static void AwesomeSort(Vector<Integer> m, int n) { // Create three vectors Vector<Integer> v1 = new Vector<>(); Vector<Integer> v2 = new Vector<>(); Vector<Integer> v3 = new Vector<>(); int i; // Traverse through the elements // of the array for (i = 0; i < n; i++) { // If elements are even and // divisible by 10 if (m.get(i) % 10 == 0) v1.add(m.get(i)); // If number is even but not divisible // by 5 else if (m.get(i) % 2 == 0) v2.add(m.get(i)); else // If number is odd v3.add(m.get(i)); } // Sort v1 in descending orde Collections.sort(v1, Collections.reverseOrder()); for (int ii = 0; ii < v1.size(); ii++) { System.out.print(v1.get(ii) + " "); } for (int ii = v2.size()-1; ii >= 0; ii--) { System.out.print(v2.get(ii) + " "); } for (int ii = 0; ii < v3.size(); ii++) { System.out.print(v3.get(ii) + " "); } } // Driver code public static void main(String[] args) { // Given Input Vector<Integer> arr = new Vector<>(); arr.add(5); arr.add(10); arr.add(30); arr.add(7); int N = arr.size(); // FunctionCall AwesomeSort(arr, N); } } // This code is contributed by abhinavjain194 # Python program for the above approach # Function to sort array in the way # mentioned above def AwesomeSort(m, n): # Create three vectors v1, v2, v3 = [],[],[] # Traverse through the elements # of the array for i in range(n): # If elements are even and # divisible by 10 if (m[i] % 10 == 0): v1.append(m[i]) # If number is even but not divisible # by 5 elif (m[i] % 2 == 0): v2.append(m[i]) else: # If number is odd v3.append(m[i]) # Sort v1 in descending orde v1 = sorted(v1)[::-1] for i in range(len(v1)): print(v1[i], end = " ") for i in range(len(v2)-1,-1,-1): print(v2[i], end = " ") for i in range(len(v3)): print (v3[i], end = " ") # Driver Code if __name__ == '__main__': # Given Input arr = [5, 10, 30, 7] N = len(arr) # FunctionCall AwesomeSort(arr, N) # This code is contributed by mohit kumar 29.
// C# program for the above approach using System; using System.Collections.Generic; class GFG { // Function to sort array in the way // mentioned above static void AwesomeSort(List<int> m, int n) { // Create three vectors List<int> v1 = new List<int>(); List<int> v2 = new List<int>(); List<int> v3 = new List<int>(); int i; // Traverse through the elements // of the array for (i = 0; i < n; i++) { // If elements are even and // divisible by 10 if (m[i] % 10 == 0) v1.Add(m[i]); // If number is even but not divisible // by 5 else if (m[i] % 2 == 0) v2.Add(m[i]); else // If number is odd v3.Add(m[i]); } // Sort v1 in descending orde v1.Sort(); v1.Reverse(); for (int ii = 0; ii < v1.Count; ii++) { Console.Write(v1[ii] + " "); } for (int ii = 0; ii < v2.Count; ii++) { Console.Write(v2[ii] + " "); } for (int ii = 0; ii < v3.Count; ii++) { Console.Write(v3[ii] + " "); } } static void Main() { // Given Input List<int> arr = new List<int>(); arr.Add(5); arr.Add(10); arr.Add(30); arr.Add(7); int N = arr.Count; // FunctionCall AwesomeSort(arr, N); } } // This code is contributed by divyeshrabadiya07. <script> // JavaScript program for the above approach // Function to sort array in the way // mentioned above function AwesomeSort(m, n) { // Create three vectors let v1 = []; let v2 = []; let v3 = []; let i; // Traverse through the elements // of the array for (i = 0; i < n; i++) { // If elements are even and // divisible by 10 if (m[i] % 10 == 0) v1.push(m[i]); // If number is even but not divisible // by 5 else if (m[i] % 2 == 0) v2.push(m[i]); else // If number is odd v3.push(m[i]); } // Sort v1 in descending orde v1.sort((a, b) => b - a); for (let i = 0; i < v1.length; i++) { document.write(v1[i] + " "); } for (let i = 0; i < v2.length; i++) { document.write(v2[i] + " "); } for (let i = 0; i < v3.length; i++) { document.write(v3[i] + " "); } } // Driver Code // Given Input let arr = [5, 10, 30, 7]; let N = arr.length; // FunctionCall AwesomeSort(arr, N); </script> Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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